Answer :
To determine the distance from a point to a plane, we will follow the formula for the distance from a point [tex]\( (x_1, y_1, z_1) \)[/tex] to a plane given by [tex]\( ax + by + cz + d = 0 \)[/tex].
1. Interpret the plane equation: The plane equation provided is [tex]\( y + 11z = 0 \)[/tex].
- To match the standard plane equation [tex]\( ax + by + cz + d = 0 \)[/tex], we can rewrite it as:
[tex]\[ 0 \cdot x + 1 \cdot y + 11 \cdot z + 0 = 0 \][/tex]
- Thus, the coefficients are:
- [tex]\( a = 0 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = 11 \)[/tex]
- [tex]\( d = 0 \)[/tex]
2. Given point: The coordinates of the point are [tex]\( (1,2,7) \)[/tex], where:
- [tex]\( x_1 = 1 \)[/tex]
- [tex]\( y_1 = 2 \)[/tex]
- [tex]\( z_1 = 7 \)[/tex]
3. Substitute the point and the coefficients into the distance formula:
The distance [tex]\( D \)[/tex] from the point [tex]\( (x_1, y_1, z_1) \)[/tex] to the plane [tex]\( ax + by + cz + d = 0 \)[/tex] is given by:
[tex]\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \][/tex]
4. Calculate the numerator:
[tex]\[ a \cdot x_1 + b \cdot y_1 + c \cdot z_1 + d = 0 \cdot 1 + 1 \cdot 2 + 11 \cdot 7 + 0 = 0 + 2 + 77 + 0 = 79 \][/tex]
- Thus, the numerator is [tex]\(|79|\)[/tex] which is [tex]\(79\)[/tex].
5. Calculate the denominator:
[tex]\[ \sqrt{a^2 + b^2 + c^2} = \sqrt{0^2 + 1^2 + 11^2} = \sqrt{0 + 1 + 121} = \sqrt{122} \][/tex]
- Thus, the denominator is [tex]\(\sqrt{122}\)[/tex].
6. Simplify [tex]\(\sqrt{122}\)[/tex]:
[tex]\[ \sqrt{122} \approx 11.045 \][/tex]
7. Calculate the distance:
[tex]\[ D = \frac{79}{\sqrt{122}} \approx \frac{79}{11.045} \][/tex]
[tex]\[ D \approx 7.152 \][/tex]
Therefore, the distance from the point [tex]\((1, 2, 7)\)[/tex] to the plane [tex]\(y + 11z = 0\)[/tex] is approximately [tex]\(7.152\)[/tex].
1. Interpret the plane equation: The plane equation provided is [tex]\( y + 11z = 0 \)[/tex].
- To match the standard plane equation [tex]\( ax + by + cz + d = 0 \)[/tex], we can rewrite it as:
[tex]\[ 0 \cdot x + 1 \cdot y + 11 \cdot z + 0 = 0 \][/tex]
- Thus, the coefficients are:
- [tex]\( a = 0 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = 11 \)[/tex]
- [tex]\( d = 0 \)[/tex]
2. Given point: The coordinates of the point are [tex]\( (1,2,7) \)[/tex], where:
- [tex]\( x_1 = 1 \)[/tex]
- [tex]\( y_1 = 2 \)[/tex]
- [tex]\( z_1 = 7 \)[/tex]
3. Substitute the point and the coefficients into the distance formula:
The distance [tex]\( D \)[/tex] from the point [tex]\( (x_1, y_1, z_1) \)[/tex] to the plane [tex]\( ax + by + cz + d = 0 \)[/tex] is given by:
[tex]\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \][/tex]
4. Calculate the numerator:
[tex]\[ a \cdot x_1 + b \cdot y_1 + c \cdot z_1 + d = 0 \cdot 1 + 1 \cdot 2 + 11 \cdot 7 + 0 = 0 + 2 + 77 + 0 = 79 \][/tex]
- Thus, the numerator is [tex]\(|79|\)[/tex] which is [tex]\(79\)[/tex].
5. Calculate the denominator:
[tex]\[ \sqrt{a^2 + b^2 + c^2} = \sqrt{0^2 + 1^2 + 11^2} = \sqrt{0 + 1 + 121} = \sqrt{122} \][/tex]
- Thus, the denominator is [tex]\(\sqrt{122}\)[/tex].
6. Simplify [tex]\(\sqrt{122}\)[/tex]:
[tex]\[ \sqrt{122} \approx 11.045 \][/tex]
7. Calculate the distance:
[tex]\[ D = \frac{79}{\sqrt{122}} \approx \frac{79}{11.045} \][/tex]
[tex]\[ D \approx 7.152 \][/tex]
Therefore, the distance from the point [tex]\((1, 2, 7)\)[/tex] to the plane [tex]\(y + 11z = 0\)[/tex] is approximately [tex]\(7.152\)[/tex].
